Find minimum value of the function $f(x,y,z)=x+y+z$ on the ball $x^2+y^2+z^2=a^2$, where $x,y,z \ge 0$ and $a>0.$
What I did:
Lagrange function:
$L(x,y,z)=x+y+z+\lambda(x^2+y^2+z^2-a^2)$.
$L_x=1+2\lambda x = 0.$
$L_y=1+2\lambda y = 0.$
$L_z=1+2\lambda z=0.$
$L_{\lambda}=x^2+y^2+z^2=a^2$.
From the first three equations: $x=y=z=-\frac{1}{2\lambda}$
From the last equation:
$\frac{3}{4\lambda^2}=a^2 \Longrightarrow\lambda=\pm \frac{\sqrt{3}}{2a}$.
Since we want $x,y,z\ge0.$
$x=y=z=\frac{a}{\sqrt{3}}$ (Took the negative $\lambda$).
And substituting back into $f(x,y,z)=\frac{3a}{\sqrt{3}}=\sqrt{3}a$.
But the answer is $a$.
Would appreciate any help, I can't find where did I make a mistake, and I don't want to assume that the final answer given is false without being sure about it, because I know I make alot of mistakes.
Thanks in advance!
What you found is the maximum of $f$ (more precisely, you found a critical point and this critical point is the maximum point of $f$) and not the minimum of $f$.
If you use Lagrange multipliers to find the minimum, you should study several cases: $(x,y,z) \in (\Bbb R^+)^{3}$, $(x,y,z)$ on the border of $(\Bbb R^+)^{3}$ (this means for example $(x,y,z= (0,y,z)$,...). And the minimum point lies on the border of $(\Bbb R^+)^{3}$
But there exists a much simpler solution: we notice that $x(a-x) \ge 0$, then
$$x(a-x) + y(a-y)+z(a-z) \ge 0 $$ $$\implies a(x+y+z) \ge x^2+y^2+z^2 = a^2$$ $$\implies x+y+z \ge a$$
The equality occurs if and only if $(x,y,z) = (0,0,a), (0,a,0)$ or $(a,0,0)$
Q.E.D